Dynamic Games

Context

• Strategic interactions and social dilemmas can sometimes be understood in terms of one-shot sandbox cases. In such settings, where time is neglected, promises, threats, and reputation play no role.

• However, players have many additional strategies available whenever there is a future, and they can use promises and threats to achieve very different outcomes compared to the atemporal cases.
  • How does time affect the outcomes of social interactions?
  • Why is reputation important whenever time is involved?
  • How can players incorporate time into their strategies?

Course Structure Overview

Lecture Structure and Learning Objectives

Structure

• Buying an Empire (Case Study)
• Repeated Games
• Sequential Games
• Backward Induction and Subgame Perfect Equilibria
• Current Field Developments

Learning Objectives

• Illustrate the relevance of time in social interactions.
• Explain how incorporating time can affect the results of interactions.
• Describe the role of backward induction when analyzing dynamic interactions.
• Illustrate the concept of subgame perfect equilibrium in dynamic settings.
• Explain how move order and patience affect the bargaining power of interacting agents.

Buying an Empire

• Auctions are typical examples of economic interactions studied by game theory.

• One of the most significant historical auctions took place in 193 AD.
• The auction for the throne of an empire!

The Praetorian Guard

The Palace Intrigues

• Commodus was the Roman emperor from 176 to 192 AD.

• He was assassinated in 192 AD by a wrestler in a bath.
• The assassination conspiracy was directed by one of his most trusted Praetorians, Quintus Aemilius Laetus.
• Pertinax, who was also a conspirator, ascended to power and paid the Praetorian guards \(3000\) denarii premium.

The Roman Empire up to Auction

• Pertinax was assassinated three months later by a group of guards!

• The Praetorian guards put the empire up to auction, taking place at the guards' camp.
• The throne was to be given to the one who paid the highest price.
• Titus Flavius Claudius Sulpicianus, Pertinax’s father-in-law, made offers for the throne (in the camp).
• Didius Julianus arrived at the camp finding the entrance barred.
• He started shouting out offers to the guards.

Hail to the Highest Bidder

But what evil have I done?

• Julianus ruled for \(66\) days!

• He was very unpopular with the Roman public because he acquired the throne by paying instead of conquering it.
• He was killed in the palace… by a soldier.
• His last words were: “But what evil have I done? Whom have I killed?”

Repeated Games

• Repeated games are games played on a finite or infinite number of dates, and at each date, the same strategic interaction is repeated.
• The repeated strategic interaction at each date is called a stage game.
• When there is a future, threats and promises can be used to sustain cooperation.
• Nevertheless, this can only happen if threats and promises are credible.

Prisoner’s Dilemma

• Consider the static prisoner’s dilemma again.
• There is a unique Nash equilibrium in which players do not coordinate (i.e., they both confess).
• Can this change if the game is repeated?

Repeated Prisoner’s Dilemma

• Suppose that the static prisoner’s dilemma is used as the stage game of a repeated game with an infinite time horizon.
• The payoff of all future dates is discounted by \(0 <\delta < 1\).

Non Coordinating Equilibrium

• One equilibrium is to play the stage game’s Nash equilibrium at each date.
• Player \(i\) gets a payoff of \(-3\) at each date, so

\[u_{n,i} = \sum_{t=0}^{\infty} \delta^{t}(-3)=-\frac{3}{1-\delta}\]

Trigger Strategies

• A trigger strategy is a strategy, where at each iteration of a repeated game an action is selected based on the coordination state of the game. If all players coordinated in the past then a coordinating action is chosen for the current iteration. Instead, past defections trigger players to choose punishing actions for the current iteration.
• Trigger strategies can be further specified as
  • Grim trigger strategies, where the punishment continues indefinitely after a player defects.
  • Tit for tat strategies, where the punishment is only applied for a limited number of dates after a defection.

Coordination in Prisoner’s Dilemma

• Players can coordinate by using trigger strategies.
• If at all previous dates the other player has denied, then deny. Otherwise, confess.
• If players coordinate, then each gets a payoff of \(-1\) at each date, thus

\[u_{c,i} = -\frac{1}{1 - \delta}\]

Can Coordination be Supported?

• If player \(i\) deviates at the current date, then her payoff at the current date is \(0\).
• At every subsequent date, her payoff is \(-3\), because her past deviation triggers the other player’s punishment.
• Therefore,

\[u_{d,i} = -3\frac{\delta}{1 - \delta}\]

• Coordination can be supported if such a deviation is not profitable, i.e.

\[u_{c,i} \ge u_{d,i} \iff \delta \ge \frac{1}{3}\]

• As long as players are patient enough, the underlying threats of trigger strategies make coordination feasible.

Sequential Games

• A game is called sequential if its players play sequentially instead of simultaneously.
• Nash equilibria also exist in such games.
• We can find some of them using backward induction.
  • The best action of the player that acts at the last date is calculated.
  • Given this best response, the best action of the player that acts at the previous to last date is calculated.
  • We continue in this fashion until we have calculated the best action of the player who acts at the initial date.

Backward Induction

• \(SPE = \left\{ \left\{Bottom, \left(Left', Right \right)\right\} \right\}\)

Subgame Perfect Equilibria

• A subgame of a dynamic game is the restriction of the game starting from a particular decision node and including all subsequent decision nodes and branches (actions) of the original game.
• A collection of strategies, one for each player, such that its restriction to each subgame of the original game constitutes a Nash equilibrium of this subgame is called subgame perfect equilibrium.
• This implies that the past does not matter in optimal decisions once a decision node is reached.
• Equilibria of this type are typically abbreviated as SPE.
• Backward induction can be used to calculate subgame perfect equilibria.

Bargaining

• A process through which two or more people decide how to share a surplus is called cooperative bargaining (or simply bargaining).
• Players negotiate how to divide the surplus (a value) in one or more rounds.

Take it or Leave it Game

• There are two players negotiating how to share a surplus of unit value.
• Player \(A\) moves first and makes an offer \(x\in[0,1]\).
• Player \(B\) moves second and decides whether to accept or reject the offer.
• If the offer is accepted, player \(A\) gets \(x\), and player \(B\) gets \(1-x\).
• If the offer is rejected, both players get zero.

First Move Advantage

• Player \(B\) accepts if \(1-x \ge 0\) or, equivalently, if \(x\le 1\).
• Player \(A\) offers \(x = 1\).

Counter-Proposal

• Suppose that player \(B\) can make a counteroffer.
• She can offer \(y\in[0,1]\) if she rejects the offer of player \(A\).
• If the counteroffer is accepted, player \(A\) gets \(1-y\), and player \(B\) gets \(y\).
• If the counteroffer is rejected, both players get zero.

Last Offer Advantage

• Player \(A\) accepts the counteroffer if \(1-y \ge 0\) or, equivalently, if \(y\le 1\).
• Player \(B\) offers \(y = 1\).
• Player \(B\) accepts the first offer if \(1-x \ge y = 1\) or , equivalently, if \(x \le 0\).
• Player \(A\) offers \(x = 0\).

Alternating Offers

• Suppose that the counter-proposal game is infinitely repeated until a deal is reached.
• Every time a player rejects an offer, she makes a counteroffer.
• Players discount every offer-round with factors \(\delta_{A},\delta_{B}\in[0,1)\).

A Recursive Equilibrium

• Suppose the game is at date \(t\) and player \(B\) makes an offer.
• Player \(A\) accepts if \(y\le 1\), so player \(B\) offers \(y=1\).
• At date \(t-1\), player \(B\) accepts if \(x\le 1-\delta_{B}\), so player \(A\) offers \(x=1-\delta_{B}\).
• At date \(t-2\), player \(A\) accepts if \(y\le 1 - \delta_{A} + \delta_{A}\delta_{B}\), so player \(B\) offers \(y=1 - \delta_{A} + \delta_{A}\delta_{B}\).
• Recursively, one can show that player \(A\) offers

\[x = \frac{1-\delta_{B}}{1-\delta_{A}\delta_{B}}.\]

The Role of Impatience

• If player \(A\) becomes more patient (\(\delta_{A}\ \uparrow\))
  • she is more willing to postpone acceptance for the next date
  • player \(B\) loses bargaining power
  • player \(A\) gets a greater part of the surplus (\(x\ \uparrow\))
• If player \(B\) becomes more patient (\(\delta_{B}\ \uparrow\))
  • she is more willing to postpone acceptance for the next date
  • player \(A\) loses bargaining power
  • player \(A\) gets a smaller part of the surplus (\(x\ \downarrow\))

Current Field Developments

• Much of modern work in game theory is dynamic.
• The folk theorem is a fundamental theoretical result stating that in infinitely repeated games, any feasible payoff vector can be achieved by a subgame perfect equilibrium if players are sufficiently patient.
• Since its development by Friedman (1971), much subsequent work has focused on equilibrium refinements that give stricter predictions.
• Another active area of work in game theory focuses on extensions of preferences of players that include behavioral traits (e.g., regret, cognitive costs, etc.).

Concise Summary

• Time is of the essence in strategic interactions.
• Many outcomes of simultaneous interactions can be overturned when time is taken into account.
• If there is a future, coordination can be supported even in cases when the corresponding simultaneous interaction results in non coordination (i.e. the prisoners' dilemma).
  • Patience is central in determining what types of coordination can be achieved and how the coordination gains can be split.
• Players can use strategies involving threats and promises to induce coordination.
  • Nevertheless, these promises and threats have to be credible.
  • The subgame perfect equilibrium is a refinement of the Nash equilibrium accounting for credibility issues.

Further Reading

• Watson (2008, chaps. 2, 14, 15)
• Belleflamme and Peitz (2010, sec. A.2)
• Varian (2010, secs. 29.5, 29.7)

Mathematical Details

The Repeated Prisoners' Dilemma

How does the strategic thinking of the two suspects change when they repeat their interaction multiple times? Repeating the game multiple times allows players to use more complicated strategies compared to the one-shot case, and the question that naturally arises is whether these strategies can be part of Nash equilibria.

Repeating the Prisoners' Dilemma Twice

To answer this question, we can start by focusing on the most straightforward possible repetition extension, namely, playing the prisoners' dilemma twice. We consider two dates \(t=0,1\), where \(0\) represents the current interaction’s date, and \(1\) represents the future interaction’s date. To examine the suspects' optimal behavior, we begin with their choices at date \(t=1\) and move backward in time.

At \(t=1\), no future interactions are remaining to be played. From the perspective of the suspects at \(t=1\), the remaining interaction is exactly equivalent to the interaction of a one-shot prisoners' dilemma game. Hence, the only Nash equilibrium of the game from the last date’s perspective is \(\{Confess, Confess\}\). The past actions cannot affect this result because they cannot affect the suspects' payoffs at \(t=1\). Thus, any equilibrium of the repeated game should involve the suspects choosing \(\{Confess, Confess\}\) at \(t=1\) (why?).

Having resolved the choices at \(t=1\), we can think of what would be reasonable to be involved in equilibrium at date \(t=0\). Given that the suspects choose \(\{Confess, Confess\}\) at \(t=1\), they receive payoffs \((-3,-3)\) at \(t=1\). Therefore, given that they both foresee this outcome, they perceive the game played at \(t=0\) as a one-shot prisoners' dilemma game where all the outcomes are translated by \(-3\) (see the next table).

Hence, the Nash equilibrium at \(t=0\) should involve the suspects choosing \(\{Confess, Confess\}\).

Describing the Nash equilibria in the repeated games requires that we bring the suspects' actions on the two dates together. Each suspect has four different strategies at her disposal, namely all combinations of the two actions available at each date. These strategies are

\begin{align*} s_{1} &= (Confess, Confess) \\ s_{2} &= (Confess, Deny) \\ s_{3} &= (Deny, Confess) \\ s_{4} &= (Deny, Deny) . \end{align*}

We have seen that any Nash equilibrium should involve both suspects choosing \(Confess\) at both dates, i.e., both suspects choosing \(s_{1}\). Therefore, the only Nash equilibrium of the repeated game is \[ \left\{(Confess, Confess), (Confess, Confess)\right\}. \] In this equilibrium, each suspect receives a payoff of \(-6\).

Repeating the Prisoners' Dilemma a Finite Number of Times

Repeating twice does not allow suspects to coordinate themselves out of always confessing. Is coordination feasible if we repeat the prisoners' dilemma more than twice? No. Our analysis in the case of 2-repetitions does not rely on the fact that the stage game (the one-shot prisoners' dilemma) is played twice, so our arguments can be recursively applied when we repeat the stage game more times.

Suppose that we repeat the stage game \(n\) times. At \(t=n-1\) (the last date), no future dates can affect how the suspects play. Thus, as in the case of \(n=2\), we conclude that any Nash equilibrium of the repeated game should involve the suspects choosing \(\{Confess, Confess\}\) at the last date.

Given that both suspects can deduce the equilibrium outcomes at date \(t=n-1\), the remaining game from the perspective of \(t=n-2\) reduces to a one-shot prisoners' dilemma game where all the outcomes are translated by \(-3\) (similarly to date \(0\) for the two repetition case; see the previous table). Thus, the Nash equilibrium at \(t=n-2\) should again involve the suspects choosing \(\{Confess, Confess\}\).

Continuing recursively in this manner, we conclude that the only Nash equilibrium in the repeated game is \[ \left\{\underbrace{(Confess, \dots, Confess)}_{n\text{ times}}, \underbrace{(Confess, \dots, Confess)}_{n\text{ times}}\right\}. \] In this equilibrium, each suspect receives a payoff of \(-3\) at each date, resulting in a total payoff \[ V_{i} = \underbrace{(-3) + \dots + (-3) }_{n\text{ times}} = -3n .\]

Repeating the Prisoners' Dilemma an Infinite Number of Times

Repeating the prisoners' dilemma any finite number of times does not allow suspects to coordinate themselves out of always confessing. The finality of the repeated interaction prevents suspects from using coordinating strategies. However, the situation becomes fundamentally different in an infinitely repeated interaction.

How can we think of strategies in an infinitely repeated interaction? Strategies are complete game plans, so they should specify which action is chosen at each eventual future date and situation. For example,

\begin{align*} s_{1} &= \{\text{Always play } C\} = ( \overbrace{C}^{0}, \overbrace{C}^{1}, \dots, \overbrace{C}^{n}, \overbrace{C}^{n+1}, \dots ) \\ s_{2} &= \{\text{Start with } C \text{ and alternate}\} = ( C, D, C, D, \dots ) \\ s_{3} &= \left\{\begin{aligned} &\text{Start with $D$ and choose $D$ if the opponent} \\ &\text{chose $D$ in all past dates; otherwise choose $C$} \end{aligned}\right\} \\ s_{4} &= \left\{\begin{aligned} &\text{Start with $D$ and choose $C$ if the opponent} \\ &\text{chose $C$ in the last date; otherwise choose $D$}, \end{aligned}\right\} \end{align*}

where we have used \(C\) to denote \(Confess\) and \(D\) to denote \(Deny\) for brevity. Strategy \(s_{3}\) is known as the grim trigger strategy, and strategy \(s_{4}\) is known as tit for tat strategy.

• Discounting Future Payoffs

  The payoffs received at the current date (\(t=0\)) are not directly comparable to the payoffs received at any future dates when suspects have a time preference for receiving rewards as early as possible. In such cases, suspects have to discount future values to make the payoffs comparable and evaluate various strategies involving very different future payoff streams. One part of the payoff is lost when discounting because a future payoff is not as valuable as a payoff with the same face value received today.

  Let \(\delta\in[0,1]\) denote the discount factor of the suspects. Receiving a unit payoff \(u_{1}=1\) at date \(t=1\) (next date) has the same value as receiving a payoff \(\delta u_{1}\) today. For example, if \(\delta=0.9\), then a unit at the next date equals to \(0.9\times 1=0.9\) units today. Similarly, receiving a unit \(u_{2}=2\) at date \(t=2\) has the same value as \(\delta u_{2}\) at date \(t=1\), and as \(\delta (\delta u_{2}) = \delta^{2} u_{2}\) at date \(t=0\). For example, if \(\delta=0.9\), then a unit in two dates from now equals \(0.9^{2}\times 1=0.81\) units today.

  The two extreme cases are \(\delta=0\) (full discounting), in which the suspects completely disregard any future payoffs, and \(\delta=1\) (no discounting), in which the suspects consider future payoffs exactly as valuable as current payoffs. We can think of the \(n\) times finitely repeated prisoners' dilemma we have previously studied as a special case of the infinitely repeated game, in which the discount factor is equal to one for the first \(n\) dates of the game, and then it becomes zero.

• Evaluating Outcomes in the Infinitely Repeated Game

  In the general case, the suspects receive an infinite stream (formally a sequence) of payoffs, one for each date, i.e., \[ \left(u_{0}, u_{1}, u_{2}, \dots, u_{n}, u_{n+1}, \dots \right), \] they discount future values using their discount factors \[ \left(u_{0}, \delta u_{1}, \delta^{2} u_{2}, \dots, \delta^{n} u_{n}, \delta^{n+1} u_{n+1}, \dots \right), \] and then they sum everything up to get an evaluation of the result,

  \begin{align*} V_{i} &= u_{0} + \delta u_{1} + \delta^{2} u_{2} + \dots + \delta^{n} u_{n} + \delta^{n+1} u_{n+1} + \dots \\ &= \sum_{t=0}^{\infty} \delta^{t} u_{t}. \end{align*}

  The utility \(V_{i}\) represents the suspect \(i\)’s evaluation of an infinitely repeated game’s outcome. The infinite sum (formally the series) of \(V\) is well defined when the payoffs \(u_{t}\) do not grow indefinitely (formally, they are bounded), and the discounting factor is strictly less than one (\(\delta<1\)).

  In repeated games, the expression of \(V_{i}\) can often be greatly simplified because the payoffs received by the suspects remain constant in many of their available strategies. For example, if both suspects choose strategy \(s_{1}\) (i.e., always \(Confess\)), then they get a payoff \(u_{t}=-3\) at all dates. In such cases, when \(u_{t}=\hat u\) for all dates, we have \[ V_{i} = \sum_{t=0}^{\infty} \delta^{t} u_{t} = \hat u \sum_{t=0}^{\infty} \delta^{t}, \] and from the geometric series result (for any \(0 \le \delta < 1\)) \[ \sum_{t=0}^{\infty} \delta^{t} = \frac{1}{1-\delta}, \] we conclude \[ V_{i} = \frac{\hat u}{1-\delta}. \]

• Equilibria with Grim Trigger Strategies

  In contrast to the case of finite repetition, at every date of the infinitely repeated prisoners' dilemma, there is always a future that the suspects can leverage for making promises and threats. This allows the suspects to achieve Nash equilibria using coordinating strategies that are impossible with finite repetitions.

  Consider the scenario where both suspects choose the grim trigger strategy \(s_{3}\), i.e., \((s_{3}, s_{3})\). Can this profile of strategies be a Nash equilibrium? Because the game is symmetric, we focus on one of the suspects. To answer this question, it suffices to (i) calculate the payoff \(V_{A}^{g}\) received by suspect \(A\) under the profile \((s_{3}, s_{3})\), (ii) calculate the payoff \(V_{A}^{d}\) received by suspect \(A\) under an alternative profile \((s', s_{3})\), where \(s'\) is the most profitable deviation of \(A\), and (iii) compare \(V_{A}^{g}\) with \(V_{A}^{d}\). If we find configurations (i.e., values for \(\delta\)) for which \(V_{A}^{g}\) is greater than the most profitable deviation \(V_{A}^{d}\), then we conclude that \(V_{A}^{g}\) maximizes the payoff of suspect \(A\) when suspect \(B\) chooses the grim trigger strategy, and consequently, due to symmetry, the profile \((s_{3}, s_{3})\) is a Nash equilibrium.

  If suspect \(A\) does not deviate from the grim trigger strategy, then both suspects end up choosing \(Deny\) at all dates, therefore, they receive a payoff of \(-1\) at every date and \[ V_{A}^{g} = \frac{-1}{1 - \delta}. \]

  If suspect \(A\) decides to deviate, any gains she obtains from deviating are the greatest at date \(t=0\) (current date) because any future gains are discounted by \(\delta\). So we can focus on the best deviation suspect \(A\) has at date \(t=0\). Given that suspect \(B\) plays the grim trigger strategy, he chooses \(Deny\) at \(t=0\), so the best deviation of \(A\) is to choose \(Confess\) and receive a zero payoff instead of \(-1\) at \(t=0\). Since suspect \(B\) plays the grim trigger strategy, he chooses \(Confess\) at all future dates \(t\ge1\) after the deviation of \(A\) at \(t=0\). Hence, the best option for suspect \(A\) is to choose \(Confess\) at all dates \(t\ge1\), receiving a payoff of \(-3\). In summary, suspect \(A\) gets

  \begin{align*} V_{A}^{d} &= 0 + \delta (-3) + \delta^{2} (-3) + \delta^{3} (-3) + \dots \\ &= -3\delta\left[ 1 + \delta + \delta^{2} + \dots \right] \\ &= -3\delta \sum_{t=0}^{\infty} \delta^{t} \\ &= \frac{-3\delta}{1-\delta}. \end{align*}

  Comparing the two evaluations, we find that \(V_{A}^{g} \ge V_{A}^{d}\) if and only if \(\delta\ge1/3\). Thus as long as the two suspects are patient enough, grim trigger strategies can be used to achieve coordination in the infinitely repeated prisoners' dilemma.

Exercises

Group A

1. Consider a two-player game in which, at date \(1\), player \(A\) selects a number \(x>0\), and player \(B\) observes it. At date \(2\), simultaneously and independently, player \(A\) selects a number \(y_{A}\) and player \(B\) selects a number \(y_{B}\). The payoff functions of the two players are \(u_{A} = y_{A} y_{B} + x y_{A} - y_{A}^{2} - \frac{x^{3}}{3}\) and \(u_{B} = -( y_{A} - y_{B} )^{2}\). Find all the subgame perfect Nash equilibria.

   We find the subgame perfect equilibrium using backward induction. At the last date, player \(B\) solves \[\max_{y_{B}} \left\{-(y_{A}-y_{B})^{2}\right\}\] and gets the best response \(y_{B}(y_{A})=y_{A}\). At the same date, player \(A\) solves \[\max_{y_{A}} \left\{y_{A}y_{B} + x y_{A} - y_{A}^{2} - \frac{x^{3}}{3}\right\}\] to get the best response \(y_{A}(y_{B};x)=x\). Solving the system of best responses, we find that \(y_{A}^{\ast}=y_{B}^{\ast}=x\) in equilibrium.

   At the initial date, player \(A\) anticipates the behavior of date \(2\) and maximizes \[\max_{x} \left\{ x^{2} - \frac{x^{3}}{3}\right\}.\] This gives the first order condition \(2x - x^{2}=0\), from which we get the solution \(x = 2\).

   Consequently, we have a unique Nash equilibrium given by \(\left\{(x=2,y_{A}=2), y_{B}=2\right\}\).

2. Find the subgame perfect equilibria of the following game. How many subgames does the game have?

   There are two subgames. The subgame perfect equilibrium is \(\left\{(U,E,H),(B,C)\right\}\).

Group B

1. Consider the alternating offer bargaining game with two players \(A\) and \(B\) negotiating how to split surplus with a unitary value. Player \(A\) moves first and makes an offer \(x\in[0,1]\). Player \(B\) moves second and decides whether to accept or reject the offer. If the offer is accepted, player \(A\) gets \(x\), and player \(B\) gets \(1-x\). If the offer is rejected, player \(B\) makes a counteroffer \(y\in[0,1]\). If the counteroffer is accepted, player \(A\) gets \(1-y\), and player \(B\) gets \(y\). If the counteroffer is rejected, the game continues with player \(A\) making an offer. Players discount every offer-round with factors \(\delta_{A},\delta_{B}\in[0,1)\).

   1. Calculate the recursive subgame perfect equilibrium.
   2. Show that if the discount factor of player \(A\) increases, then she can get a greater part of the surplus.
   3. Show that if the discount factor of player \(B\) increases, then player \(A\) gets a smaller part of the surplus.

   1. Suppose that we are at time \(t\) and player \(B\) makes an offer. Player \(A\) accepts the offer if and only if \(1-y \ge 0\). Thus, player \(B\), aiming at maximizing her share, solves \[\max_{1-y \ge 0} y\] and offers \(y=1\). At time \(t-1\), player \(B\) knows that if the game proceeds to time \(t\), player \(A\) will accept the offer \(y=1\). Therefore, if she waits for the next time point, player \(B\) can gain \(\delta_{B}\cdot 1\) in present value terms. Hence, she accepts any offer of player \(A\) securing her at least this gain, namely \(1-x \ge \delta_{B}\). Player \(A\) anticipates this behavior, solves \[\max_{1-x \ge \delta_{B}} x,\] and offers \(x=1-\delta_{B}\).

      At time \(t-2\), player \(A\) knows that if she delays, she can gain \(1-\delta_{B}\), which corresponds to gaining \(\delta_{A}(1-\delta_{B})\) discounted in time \(t-2\) terms. She, therefore, accepts an offer if \(1-y \ge \delta_{A} - \delta_{A}\delta_{B}\). Thus, player \(B\) solves \[\max_{1-y \ge \delta_{A} - \delta_{A}\delta_{B}} y\] and offers \(y = 1 - \delta_{A} + \delta_{A}\delta_{B}\). In a similar manner, one argues that at time \(t-3\) player \(A\) solves \[\max_{1-x \ge \delta_{B} - \delta_{A}\delta_{B} + \delta_{A}\delta_{B}^2} x,\] and offers \(x=1 - \delta_{B} + \delta_{A}\delta_{B} - \delta_{A}\delta_{B}^2\).

      Repeating the aforementioned recursive arguments, we find that player \(A\)’s offer at date \(t-(2n-1)\) (observe that, because we started with player \(B\), player \(A\) moves at odd offer-rounds) is

      \begin{align*} x &= 1 - \delta_{B} + \delta_{A}\delta_{B} - \delta_{A}\delta_{B}^2 + \dots + \delta_{A}^{n-1}\delta_{B}^{n-1} - \delta_{A}^{n-1}\delta_{B}^{n} \\ &= 1 + \dots + \delta_{A}^{n-1}\delta_{B}^{n-1} - \delta_{B}\left(1 + \dots + \delta_{A}^{n-1}\delta_{B}^{n-1}\right) \\ &= \left(1 - \delta_{B}\right) \sum_{j=0}^{n-1} \delta_{A}^{j}\delta_{B}^{j}. \end{align*}

      Letting \(n\) go to infinity, the share of player \(A\) converges to

      \begin{align*} x = \frac{1 - \delta_{B}}{1 - \delta_{A}\delta_{B}}. \end{align*}

   2. We can find the effect of \(\delta_{A}\) on player \(A\)’s share by calculating

      \begin{align*} \frac{\partial x}{\partial \delta_{A}} = \delta_{B}\frac{1 - \delta_{B}}{\left(1 - \delta_{A}\delta_{B}\right)^{2}}, \end{align*}

      which is positive.

   3. From

      \begin{align*} \frac{\partial x}{\partial \delta_{B}} = - \frac{1 - \delta_{A}}{\left(1 - \delta_{A}\delta_{B}\right)^{2}} < 0, \end{align*}

      we see that player \(A\)’s share is decreasing in \(\delta_{B}\).

References

References

Topic's Concepts

• bargaining
• cooperative bargaining
• subgame perfect equilibrium
• subgame
• sequential
• tit for tat strategies
• grim trigger strategies
• trigger strategy
• stage game
• repeated games